Hey guys,

I have a final on Thursday and I'm struggling with about 12 problems. I'd appreciate it if someone who is a pro at calculus can PM me and I can send you the problems via screenshot. I just need you to write the steps to solving it. I know the answer to the problem, I need to find out how to get to that answer.

Thanks!

This post was edited by Dune1 on Apr 25 2015 02:55pm

post here ?

1)

x² - 5x - 24 = ( x - 8 ).( x + 3 )

hence, your fraction simplify as : x + 3, as soon as x is not 8.

Now you can study limit, as x tends to 8, while x < 8, of x + 3 : the limit (to the left) is 11,
and similarly, the limit to the right ( x > 8 ) is 11 as well.

You can conclude that the limit exists, and is 11.

2)

N(4) - N(3) = 20 - 55 = -35

N'(p) = - 10p
N'(3) = -30
N'(4) = -40

3)

f(x) = 8.( 5x² + 9 )^0.5

f '(x) = 8*0.5*10x / ( 5x² + 9 )^0.5
f '(x) = 40x / ( 5x² + 9)^0.5

4)

for every x non-zero :
f '(x) = - (8/9) * x ^ (8/9 - 1)
f '(x) = - (8/9) * x ^ (-1/9)

f ''(x) = - (8/9) * (-1/9) * x (-1/9 - 1)
f ''(x) = (8/81) * x ^ (-10/9)

f ''(x) doesn't exist, indeed ; and you got right as well for f ''(1) and f ''(-1).

5)

f '(x) = 9x² - 6x - 3 = 3(3x² - 2x - 1) = 3( x - 1 ).(3x + 1 )
Since f is defined over [-1;0], since -1/3 is the only root of f '(x) in that interval :
f meets its maximal value at -1/3,
and to know the minimum value of f, you'll have to compare f(-1) and f(0) : f(-1) = 4 and f(0) = 7.

6)

f '(x) = (2/3)* 2x * (x² + 12) ^ (-1/3)

f '(x) = 0 if and only if x = 0,
f '(x) > 0 if and only if x > 0,
f '(x) < 0 if and only if x < 0.

f is increasing on [0;3] and decreasing on [-5;0].
The absolute minimum occurs at x=0, and you have to compare f(-5) and f(3) : this is quite obvious since f is even.

7)

elasticity is not mathematics.


8)

P '(x) = -3x² + 18x + 120 = 3 ( -x² + 6x + 40) = 3 ( x - 10 ).( -x - 4 )
Given the interval [5 ; +∞[ , the maximum value of P occurs at x = 10.
P(10) = 800 -> $ 800,000
x = 10 -> 1,000,000 pillows

9)

correct, not much to say

10)

You are asked for an anti-derivative of P '(x), which is P(x) = (25/2) x^4 + 20 x^3 + C
and since P(0) = -20, C = -20.

For 300 pounds, replace x by 3. Answer is correct.

11)

Let u = t^7 + 5t
du = ( 7t^6 + 5).dt

√(t^7 + 5t).(7t^6 + 5).dt = √u.du

∫√u.du = (2/3).u√u = (2/3).(t^7 + 5t).√(t^7 + 5t) = (2/3).(t^7 + 5t)^(3/2)

+ C obviously

12)

I = [ - exp(-0.3A) + 4 ln(A) ] = - exp (-0.9) + exp (-0.6) + 4 ln(3) - 4 ln(2)
I ~ 1.764

13)

Uneasy to see, but you could be right.


14)

Correct.

15)

f(x) = 4 - x² is an even function.

∫ f between 0 and 2 is 8 - 8/3 = 16/3
∫ f between 2 and 3 is (12 - 27/3) - 16/3 = - 7/3 -> shaded region worth 7/3

2 * ( 16/3 + 7/3 ) = 2*23/3 = 46/3

16)

answer a) because : the dashed line start just over 1,000 and is decreasing, and the solid line start at 0 and is increasing.

Solve for q : S(q) = D(q)
q² + 8q = 1020 - 18q - q²
2q² + 26q - 1020 = 0
q² + 13q - 510 = 0
(q - 17).(q + 30) = 0
q = 17 (since -30 is out of range)

I don't get what consumer/producer's surplus mean.

Still not understanding these problems. Can someone break them down step-by-step?

http://i.imgur.com/3o2QQ11.png
http://i.imgur.com/KGsWKEf.png
http://i.imgur.com/utzbO0N.png

This post was edited by Dune1 on Apr 29 2015 08:48am

////

This post was edited by saber_x3 on Apr 29 2015 09:32am

I need some more help, specifically with word problems. I would love to find someone with AIM that can help me break these problems down through that. Quicker and more efficient that way.